Structural Design Project

CAMPBELL HALL

UNIVERSITY OF CALIFORNIA, BERKELEY

CEE 123 L: Structural Concrete Design Project

Professor J. Moehle

ALEXANDER BILL MELISSA MEIKLE

ANDREW RICHARD QUDSIA WAHAB

MAY 9, 2014

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First Floor

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East elevation

North elevation

TABLE OF CONTENTS Transmittal Letter

Cost Estimation

Design Calculations

General Project Description

Structural system

Building code and references

Loading criteria

Gravity loads and assumed unit weights

Wind loading criteria

Seismic loading criteria

Materials

Load combinations

Analysis and design approach summary

Geotechnical Information

Gravity system design

Slab design

Beam design

Girder design

Column design

Lateral system design

Seismic load calculations

Shear wall design

Wind load calculations

Retaining wall design

Foundation design

Spread footing design

Strip footing design

Structural Drawings

Layout and General Notes

First Floor Framing Plan

Building Elevation

Foundation Plan

Footing Detail

Slab, Beam, and Girder Detail

Beam and Girder Detail

Column Detail

Shear Wall and Retaining Wall Detail

Roof Framing Plan

Basement Framing Plan

Diaphragm Plan

Appendix A

CalStructures

Davis Hall

University of California

Berkeley, CA 94720-1712

Jack P Moehle

775 Davis Hall

University of California

Berkeley, CA 94720-1712

Dear Professor Moehle,

Please find attached our submission of our proposal for the Campbell Hall Replacement Project.

The purpose of this proposal is to present you with our unique design solution, along with initial

cost estimate, for this prestigious project. The report was assigned on January 24th 2014. The

fee we are proposing for this project is $2,627,489.33, which we believe is great value for the

design we present in this proposal.

The report includes a brief overview of the project and the specifications provided by the client

and how we have met and exceeded them. We then move into the structural design for this

reinforced concrete solution. We have analyzed our solution for wind, gravity, earth and

earthquake loading ensuring that every scenario is covered and a perfectly functioning building

is achieved. We have included detailed design of the elevated slabs, ground bearing slab,

foundations, retaining walls, shear walls, columns and the building diaphragm. We have also

produced and included detailed construction documents with the detailing of each member

displayed clearly in the attached drawings. A cost estimate has been formulated and is

displayed in this document along with a cost breakdown into each member of the building. All

calculations have been attached also.

We would greatly appreciate to hear your feedback on our findings and we look forward to

hearing the result of who has the honor of delivering this great project.

Yours sincerely,

Alexander Bill, CalStructures Summer Intern

Melissa Meikle, CalStructures Senior Structural Engineer

Andrew Richard, CalStructures Senior Structural Engineer

Qudsia Wahab, CalStructures Senior Structural Engineer

Campbell Hall “COST ESTIMATION” Date: 03/28/2014

Calculated by: A.B, M.M, A.R, Q.W

CE123L Group Project - Cal Structures (Group 6) - Cost Estimation - 04/27/14

Alexander Bill - Melissa Meikle - Andrew Richard - Qudsia Wahab

Grade beams and spread footings Unit Quantity Cost ($/unit) Total Cost Per Member

Formwork

Spread Footing C1 8'x8'x2.5' SF 320 7.5 $2,400.00

Spread Footing C2 12'x12'x2.5' SF 480 7.5 $3,600.00

Spread Footing C3 15'x15'x3' SF 1620 7.5 $12,150.00

Strip Footing SF 360 7.5 $2,700.00

Total 2780 $20,850.00

Reinforcement

Spread Footing C1 lb 12544 1 $12,544.00



Strip Footing lb 14700 1 $14,700.00

Total 174538 $174,538.00

Concrete & placing - 4000 psi

Spread Footing C1 CY 23.7037 200 $4,740.74

Spread Footing C2 CY 53.33333 200 $10,666.67

Spread Footing C3 CY 225 200 $45,000.00

Strip Footing CY 27.77778 200 $5,555.56

Total 329.8148 $65,962.96

Grade Beams and Spread Footings Total Cost $261,350.96

Slab-on-grade

Edge forms

Basement Slab LF 400 7 $2,800.00

Total 400 $2,800.00

Aggregate base over vapor retarder

Basement Slab SF 10000 1.5 $15,000.00

Total 10000 $15,000.00

Control joints

Basement Slab SF 10000 0.4 $4,000.00

Total 10000 $4,000.00

Reinforcement

Basement Slab lb 130666.7 1 $130,666.67

Total 130666.7 $130,666.67


Basement Slab CY 246.9136 195 $48,148.15

Total 246.9136 $48,148.15

Finish & cure

Basement Slab SF 10000 1 $10,000.00

Total 10000 $10,000.00

Slab-on-grade cost $210,614.81

Concrete Walls

Formwork

Shear Walls SF 9540 15 $143,100.00

Retaining Wall SF 4500 15 $67,500.00

Total 9540 $210,600.00

Reinforcement –LFRS

Shear Walls lb 132300 1.25 $165,375.00



Retaining Wall lb 102900 1.25 $128,625.00

Total 132300 $294,000.00


Shear Walls CY 250 210 $52,500.00

Retaining Wall CY 194.4444 210 $40,833.33

Total 250 $93,333.33

Finish & cure - Architectural

Shear Walls SF 9540 2.5 $23,850.00

Retaining Wall SF 4500 2.5 $11,250.00

Total 9540 $35,100.00

Concrete Walls - Building Total $633,033.33

Columns

Formwork

C1 4 x (16"x16") SF 960 10 $9,600.00

C2 4 x (18"x18") SF 1080 10 $10,800.00

C3 9 x (24"x24") SF 3240 10 $32,400.00

Total 5280 $52,800.00

Reinforcement

C1 lb 25088 1 $25,088.00

C2 lb 31752 1 $31,752.00

C3 lb 127008 1 $127,008.00

Total 183848 $183,848.00

Concrete & placing – 4000 psi

C1 CY 47.40741 210 $9,955.56

C2 CY 60 210 $12,600.00

C3 CY 240 210 $50,400.00

Total 347.4074 $72,955.56

Finish & cure - Architectural

C1 SF 960 2.5 $2,400.00

C2 SF 1080 2.5 $2,700.00

C3 SF 3240 2.5 $8,100.00

Total 5280 $13,200.00

Columns Total Cost $322,803.56

Suspended slabs , with or without drop panels

Form & shore slab soffit

S1 12 bays SF 16875 12.5 $210,937.50

S2 4 bays SF 6525 12.5 $81,562.50

Total 23400 $292,500.00

Form & shore beams/girders

B1 SF 1650 12.5 $20,625.00

B2 SF 4400 12.5 $55,000.00

B3 SF 3300 12.5 $41,250.00

G1 SF 1650 12.5 $20,625.00

G2 SF 3300 12.5 $41,250.00

Total 14300 $178,750.00

Reinforcement

S1 lb 294000 1 $294,000.00

S2 lb 98000 1 $98,000.00

B1 lb 11760 1 $11,760.00



B2 lb 31360 1 $31,360.00

B3 lb 23520 1 $23,520.00

G1 lb 11760 1 $11,760.00

G2 lb 23520 1 $23,520.00

Total 493920 $493,920.00

Concrete & placing – 4000 psi

S1 CY 555.5556 215 $119,444.44

S2 CY 185.1852 215 $39,814.81

B1 CY 22.22222 215 $4,777.78

B2 CY 59.25926 215 $12,740.74

B3 CY 44.44444 215 $9,555.56

G1 CY 22 215 $4,777.78

G2 CY 44 215 $9,555.56

Total 933 $200,666.67

Finish & cure top

S1 SF 22500 0.5 $11,250.00

S2 SF 7500 0.5 $3,750.00

Total 30000 $15,000.00

Finish & cure exposed soffit

S1 SF 16875 0.5 $8,437.50

S2 SF 6525 0.5 $3,262.50

B1 SF 1650 0.5 $825.00

B2 SF 4400 0.5 $2,200.00

B3 SF 3300 0.5 $1,650.00

G1 SF 1650 0.5 $825.00

G2 SF 3300 0.5 $1,650.00

Total 37700 $18,850.00

Suspended Slabs , With or Without Drop Panels Total Cost

$1,199,686.67

Total Project Cost $2,627,489.33

GENERAL PROJECT DESCRIPTION

University of California intends to construct a new 30,000 square foot building that will house

classrooms, offices, and library stacks on the U.C. Berkeley Campus. It will also have a roof garden.

Campbell Hall will be constructed of concrete and have a basement and two stories above ground. The

basement is intended for library, the first floor is intended for classrooms, and the second floor is

intended for office spaces. The seismic loading is considered in the design of the building because the

building is located 1,500 feet away from Hayward fault. The project is registered to receive a LEED

Platinum rating. The expected completion date is December 4th 2014.

PROJECT DESIGN AND ENGINEERING

Project Manager: Professor Jack Moehle

Structural Engineer: CalStructures

Geotechnical Engineer: Kutt and Filit

Architect: Mahyar Mostafavy

General Contractor: DPR Construction

STRUCTURAL SYSTEM

The building will have a system consisting of frames to resist gravity loads and structural walls to resist

wind and earthquake loads. The frames consist of beams, girders, and columns, whereas the lateral

system consists of four shear walls located in the perimeter of the building.

BUILDING CODE AND REFERENCES

The building is designed in compliance to ACI 318-11 and ASCE 07-10.

Other references:

USGS Detailed Report (http://earthquake.usgs.gov/designmaps)

NEHRP Seismic Design Technical Brief

LOADING CRITERIA

Unit weight of concrete is assumed to be 150 pcf and is used for the calculation of structural members’

self-weight such as slabs, beams, girders, and columns.

The dead and live load breakdown for the building is tabulated below:

Table 1: Load Breakdown

ROOF DEAD LOAD LIVE LOAD

Self Weight 150 pcf Garden 100 psf

HVAC 15 psf

Lighting 5 psf

http://earthquake.usgs.gov/designmaps

LEVEL2 DEAD LOAD LIVE LOAD

Self Weight 150 pcf Office 50 psf

HVAC 15 psf

Lighting 5 psf Cladding 15 psf

LEVEL1 DEAD LOAD LIVE LOAD

Self Weight 150 pcf Classes 100 psf

HVAC 15 psf Lighting 5 psf

Cladding 15 psf

BASEMENT DEAD LOAD LIVE LOAD

Self Weight 150 pcf Shelf 300 psf

Cladding 15 psf

The loading criteria for seismic loading are as follows:

Site Class: B

Risk Category: II R = 5, Importance factor = 1

Seismic Design Category: D

Deflection Amplification Factor, Cd = 5

Over-strength factor, Ωo = 2.5

Seismic weight =

Base shear, V=

Wind loads are also considered, and the loading criteria are as follows:

Exposure: B

Design wind speed = 85 mph

Gust factor = 0.85

Importance factor = 1

Topographic effect = 1

Wind directionality factor = 0.85

MATERIALS

Table 2: Material Properties

Soils Density = 110 pcf

Concrete Density = 150 pcf, f’c = 4 ksi

Reinforcement grade ASTM A775, fy = 60 ksi

LOAD COMBINATIONS

1.4 * Dead Load

1.2 * Dead Load + 1.6 * Live Load

1.0 * Dead Load + 0.25 * Live Load

ANALYSIS AND DESIGN APPROACH SUMMARY

We use LRFD method for the most part of our analysis.

SAP2000 is used to analyze and obtain design moments for slabs, beams, girders and columns.

XTRACT is used to obtain P-M interaction diagrams for columns.

After obtaining the design moment, we generally proceed by selecting the required reinforcement and

making sure that demand is less than capacity. We then perform the necessary checks as required by

ACI 318-11.

GEOTECHNICAL INFORMATION

According to geotechnical firm of Kutt and Filit, the soil bearing capacities and geotechnical design

recommendations are as follows:

Dead load only 3500 psf

Combined dead plus live load 4500 psf

Combined dead plus live plus wind or seismic load 6500 psf

Soil active pressure = 30 pcf per foot of retaining wall height, applied as a triangular load to the

wall, with 0 pressure at grade

Soil passive pressure = 50 pcf per foot of soil applied as a triangular load to the wall with 0

pressure at 18” below grade or basement slab

Soil seismic surcharge = 7 h where h is the height of retained soil, applied as a uniform load to

the wall.

Slab on grade should be a minimum of 6” thick and be built over a vapor retarder placed over a

minimum of 4” of coarse aggregate. This construction can support a live load of up to 150 psf.

Heavier loads should use a slab at least 8” thick

Campbell Hall “SLAB” Pg. 1 out of 6 Date: 02/21/2014


SLAB DESIGN

The slab is designed as a one-way sab system with a span of 12’ 6”. SAP2000 is used to find the

design moment, Mu. In SAP2000, the slab is modeled using a fixed support at the left or right

end when there is a shear wall. This resulted in a smaller design moment. Using the design

moments, the slab is analyzed a T-beam. If the analysis shows that the slab must be designed as a

T-beam, then the effective flange width, beffective, must be used in the calculations. Otherwise, the

slab can be designed using a rectangular cross-section. When the moment is positive, steel is

placed along the bottom, and when the moment is negative, steel is placed along the top of the

slab. Cracking control spacing and temperature and shrinkage reinforcement are determined for

the maximum design moments. The below calculations adhere to ACI 318-11, ASCE 7, and

ASTM reinforcement.

Assumptions

Design live load for Level 1 is 100 psf (from ASCE 7)

Maximum aggregate size = 1”

f’c = 4000 psi at 28 days

Reinforcing steel shall conform to ASTM A615 Gr. 60 or A616 Gr. 60 or A 617 Gr. 60

moment

shear

Design for Positive Moment Resisting Reinforcement on Bottom (EW Direction):

Choosing reinforcement for positive moment (bottom steel):

From SAP2000 analysis, Mu = 5.68 k-ft = 68.16 k-in

o where

fy = 60 ksi, f’c = 4 ksi

d = 8 – 0.75 – 0.5 = 6.75”

b = 12”

Selecting Bars:



Table 3.1: Bar Number and Spacing To Resist Moment on Bottom

No. Abar (in2) Spacing (in)

3 0.11 6.95

4 0.2 12.63

5 0.31 19.6

o Where spacing

o At this stage, pick No. 4 at 12” spacing

Check:

Crack control: s ≤ 15 (

) -2.5 Cc = 13.13”

≤ 12 (

) = 12” governs

o where Cc = ¾”, =

=

= 40,000

Temperature and Shrinkage requirement:

=

= 0.002 ≥ 0.0018 (OK)

for positive moment resisting reinforcement on bottom (EW direction)

Choosing Temperature and Shrinkage reinforcement:

≥ 0.0018 =

and s ≤ 5h ≤ 18”

Selecting Bars:

Table 3.2: Bar Number and Spacing For Temperature and Shrinkage


3 0.11 7.64

4 0.2 13.89

5 0.31 21.5 18 (not OK)

o Where

for Temperature and Shrinkage reinforcement (NS direction)

Design for negative moment resisting reinforcement on top (EW Direction):

Choosing reinforcement for negative moment (top steel) along the beam:

From SAP2000 analysis, Mu = - 7.34 k-ft = -88.1 k-in

Check this moment against



o b effective =

=

= 6.25’ governs

≤ bw +16 tf = 12” + 16 (8”) = 140” = 11.67’

≤ lslab = 12.5’

=

= - 1.03 k-ft < - 7.34 k-ft (from SAP)

o Hence, use Mu = -7.34 k-ft

o

o where


d = 8 – 0.75 – 0.5 = 6.75”

b = 12”

Selecting Bars:

Table 3.3: Bar Number and Spacing To Resist Moment on Top


3 0.11 5.28

4 0.2 9.6

5 0.31 14.88

b effective

2.625’

BEAM



o Where

o At this stage, pick No.5 at 12” spacing

Check:

Crack control: s ≤ 15 (

) -2.5 Cc = 13.13”

≤ 12 (

) = 12” governs

o where Cc = ¾”, =

=

= 40,000

Temperature and Shrinkage requirement:

=

= 0.003 ≥ 0.0018 (OK)

for negative moment resisting reinforcement on top (EW direction)

Design for reinforcement of top steel along girder to prevent crack (NS Direction):

Choosing reinforcement for top steel along the girder to prevent crack:

o b effective =

=

= 6.25’ governs

≤ bw +16 tf = 12” + 16 (8”) = 140” = 11.67’

≤ lslab = 12.5’

b effective

2.625’

GIRDER



=

= - 1.03 k-ft

o Hence, use Mu = -7.34 k-ft

o

o where


d = 8 – 0.75 – 0.5 = 6.75”

b = 12”

Selecting Bars:

Table 3.4: Bar Number and Spacing To Prevent Crack


3 0.11 39

for reinforcement on top along girder to prevent crack (NS direction)

Sample SAP2000 analysis (for slab):

Dead Load

Combination of Live Load producing maximum moment in slab

Moment diagram for the corresponding dead and live load using load combo 1.2D + 1.6L



SLAB SCHEDULE

Table 3.5: Slab Schedule

MARK DEPTH REINFORCEMENT

BOTTOM TOP T & S

S1 8” No. 4 @ 12”

(both continuous

and non-

continuous end)

No. 5 @ 12”

(both continuous

and non-

continuous end)

No. 4 @ 12”

Notes:

1. Place main reinforcing steel so that the bottom of steel is 2” above forms in beams and

girders, and ¾” in slabs.

2. For slab S1, the top reinforcement in the non-continuous end must be hooked into the

support, while all the bottom reinforcement in the non-continuous end must be extended

6” into the support.

3. Place #3 x 6.25’ @ 12” reinforcing steels on the top of slab (NS direction) along the

girder.

4. For both slab S1 and S2, one in every four bottom reinforcement shall be continuous all

the way.

Campbell Hall “BEAMS” Pg. 1 out of 7 Date: 02/14/2014


BEAM 1 DESIGN

The beam spans the NS direction with a span of 25’. SAP2000 is used to find the design

moment, Mu. In SAP2000, the beam is modeled using a fixed support where there are columns

and shear walls. Using the design moments, the beam is analyzed as a T-beam. If the analysis

shows that the slab must be designed as a T-beam, then the effective flange width, beffective, must

be used in the calculations. Otherwise, the beam can be designed using a rectangular cross-

section. When the moment is positive, steel is placed along the bottom, and when the moment is

negative, steel is place along the top of the slab. The moment capacity, phi factor, area of steel,

spacing for cracking, inner spacing, and stirrups are determined for the maximum design

moments. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.

Design at Support (Rectangular Cross-Section):

Given From SAP2000:

From load combinations, the maximum negative moment is 189.53 kip-ft.

Assumptions:

No. 8 bar reinforcement

No. 4 stirrup

moment

shear

Parameters:

From ACI 318-11, the coefficient β1 can be determined from the compressive strength of the

concrete.

1.5



Rectangular Design:

o

Selecting Bars:

Table 3.6: Number of Required Bars for Design a the Support

No. Abar (in2) No. Req’d As, provided (in

2)

5 0.31 7 2.17

6 0.44 5 2.20

7 0.60 4 2.40

8 0.79 3 2.37

9 1.00 3 3.00

10 1.27 2 2.54

o Choose 3 No. 8 bars

Check:

Moment:

o

o

Phi Factor:

o

o



o

o

o

As:

o As max:

o As min:

o

Spacing:

Cracking:

o

o

o

contro s

- 2 1.5

Inner Spacing:

o , or

ma agg. si e 1

- 2 1.5

Design at Midspan:

Given From SAP2000:

From load combinations, the maximum positive moment is 123.36 kip-ft.

Check for T Beam design:



If β1c> tf then design as a T-beam

As c is unknown the current design will be carried out assuming it’s a T-beam

Effective width calculation

o Parameters:

Effective width is equal to the lowest of the following:

o

=

Assumptions:

o εs > εy

o φ = 0.9

o Assuming 3 No.8 bars

o Assuming No. 4 stirrups

Depth to neutral axis, c, calculation:

Parameters:

o

o

o

o

b effective

BEAM



o

Check to see if neutral axis falls below tf:

Check assumptions:

Design for moment steel:

o

Selecting Bars:

Table 3.7: Number of Required Bars for Design at Midspan


2)

6 0.44 5 2.20

7 0.60 3 1.80

8 0.79 2 1.58

9 1.00 2 2.00


Stirrup Design:

Shear:

Are stirrups req’d?



o

o

o

o = 38.0 K

Is beam sufficient for applied shear?

o

Check:

Check Smax

o

o

10 2 contro s

Check AV,min

o

o

contro s

Check req’d Vs

o

Select stirrup:

Table 3.8: Stirrup Size and Spacing

Type Av (in2) S (in)

No. 3 0.22 16

No. 4 0.40 29

Max spacing (in) 10

Choose No. Stirrup at 10” spacing

1st stirrup is ocated at 5” from the face of support

Campbell Hall “BEAMS”

Pg.7 out of 7 Date: 02/14/2014 Calculated by: A.B, M.M, A.R, Q.W

BEAM SCHEDULE

Table 3.9: Beam Schedule

BEAM SCHEDULE

MARK SIZE BOTTOM TOP STIRRUPS

W H “A” BAR “B” BAR SIZE

NO.

SPACING

B1 12” 2 ” 2 No. 7 1 No. 7 3 No. 8

CONT.

30 No. 3 1 @ 5”, 1 @ 10”

B2 12” 2 ” 2 No. 8 2 No. 8 4 No. 9 30 No. 3 1 @ ”, 1 @ 6”

B3 12” 2 ” 2 No. 7 1 No. 7 4 No. 8 30 No. 3 1 @ .5”, 1 @ 9”

Campbell Hall “GIRDERS” Pg. 1 out of 8 Date: 02/21/2014


GIRDER 1 DESIGN

The girder spans the EW direction with a span of 25’. SAP2000 is used to find the design

moment, Mu. In SAP2000, the girder is modeled using a fixed support where there are columns

and shear walls. Using the design moments, the beam is analyzed as a T-beam. If the analysis

shows that the slab must be designed as a T-beam, then the effective flange width, beffective, must

be used in the calculations. Otherwise, the beam can be designed using a rectangular cross-

section. When the moment is positive, steel is placed along the bottom, and when the moment is

negative, steel is place along the top of the slab. The moment capacity, phi factor, area of steel,

spacing for cracking, inner spacing, and stirrups are determined for the maximum design

moments. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.

Design at Support (Rectangular Cross-Section):

Given From SAP2000:

From load combinations, the maximum negative moment is 191.091 kip-ft.

Assumptions:

No. 8 bar reinforcement

No. 4 stirrup

o ent

shear

Parameters:

.5

Rectangular Design:



o

Selecting Bars:

Table 3.10: Number of Required Bars for Design at Support


2)

5 0.31 7 2.17

6 0.44 5 2.20

7 0.60 4 2.40

8 0.79 3 2.37

9 1.00 3 3.00

10 1.27 2 2.54


Check:

Moment:

o

o

Phi Factor:

o

o

o

o

o

As:

o As max:

o As min:



o

Spacing:

Cracking:

o

o

o

contro s

Inner Spacing:

o or

a agg. si e

contro s

Design at Midspan:

Given From SAP2000:

From load combinations, the maximum positive moment is 183.68 kip-ft.

Check for T Beam design:

If β1c> tf then design as a T-beam

As c is unknown the current design will be carried out assu ing it’s a T-beam

Effective width calculation

o Parameters:

b effective

GIRDER



Effective width is equal to the lowest of the following:

o

=

Assumptions:

o εs > εy

o φ = 0.9

o Assuming 3 No.8 bars

o Assuming No. 4 stirrups

Depth to neutral axis, c, calculation:

Parameters:

o

o

o

o

o

Check to see if neutral axis falls below tf:

Check assumptions:



Design for moment steel:

o

Selecting Bars:

Table 3.11: Number of Required Bars for Design at Midspan


2)

7 0.60 4 2.40

8 0.79 3 2.37

9 1.00 2 2.00


Check:

Moment:

o

o

Phi Factor:

o

o

o

o

o

As:

o As max:

o As min:

o



Spacing:

Cracking:

o

o

o

contro s

Inner Spacing:

o or

a agg. si e

Stirrup Design:

Shear:

From SAP:

Are stirrups req’d?

o

o

o

o Since

and

stirrups are required and required for strength

Is girder sufficient for applied shear?

o

o Okay, girder is sufficient.

Check:

Check Smax

o

o

o

0 2 contro s

Check AV,min

o

o

contro s



Check req’d Vs

o

o

Select stirrup:

Table 3.12: Stirrup Size and Spacing

Type Av (in2) S (in)

No. 3 0.22 22

No. 4 0.40 40

Max spacing (in) 10

o Choose No. 3 Stirrup at 0” spacing

o 1st stirrup is ocated at 5” fro the face of support

o Stirrups can be discontinued when

SAMPLE SAP2000 ANALYSIS

Dead load is placed along the entire girder. Live load is placed along the first, third, fourth span

to produce the maximum moments.

Moment diagram for the corresponding dead and live load using load combination 1.2D + 1.6L

This generated a positive maximum moment of 183.678 k-ft and a negative maximum moment

of 191.091 k-ft.

Shear diagram for the corresponding dead and live load using load combination 1.2D+1.6L



GIRDER SCHEDULE

Table 3.13: Girder Schedule

GIRDER SCHEDULE

MARK SIZE BOTTOM TOP STIRRUPS

W H “A”

BAR

“B” BAR SIZE

NO.

SPACING

G1 12” 24” 2 No. 8 1 No. 8 3 No. 8

CONT.

30 No. 3 @ 5” @ 0”

G2 12” 24” 3 No. 8 2 No. 8 6 No. 8

(2 LAYERS)

30 No. 3 @ ” @ 6”

Campbell Hall “COLUMNS” Pg. 1 out of 5 Date: 03/07/2014


COLUMN DESIGN

The columns for the basement and first levels are designed. The column has a height of 15 feet. The

axial force is calculated from the given loads. SAP2000 is used to find the design moment, Mu. The

dimensions of the column and the amount of steel required are calculated. XTRACT is used to obtain a P-

M diagram for the column. For the required transverse steel, the probable moment strength (Mpr) is

used. The below calculations adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.

Column 1:

To calculate the axial force the on the column, the loads from the slab and the beams/girders are

considered. Calculations are done in Excel and are attached in Appendix A.

In order to calculate the maximum moment on the column, two frames are analysed. Distributed dead

load is on all spans, whereas distributed live load is only on the first and third spans. The figure shows

the models from SAP2000 that are used to determine the maximum moment on the column. Another

figure shows the moment diagram used to obtain the maximum moment.

(a) (b)

a) The location of the distributed dead load on the frame. b) The location of the distributed live load on

the frame.

The moment diagram of the first level



Ultimate Moment, Mu, and Ultimate Axial Load, Pu

Initial Values

Gross Cross Sectional Area, Ag, Calculation

Longitudinal Reinforcement Area, Ast, Calculation

Spacing Requirements

Allowable center-to-center spacing



The P-M interaction Diagram for column 1, shown in Figure 1 is obtained from XTRACT by using an initial

design for the column. A rectangular column with 8 No.8 bars is input. The type of steel is specified as A

615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain is

0.003.

Figure 1: P-M Interaction Diagram for Column1

As it can be seen in Figure 1, the point Pu, Mu is inside of the curve, which means that the

design of the column is adequate.

Column Shear

(This value is obtained from XTRACT)

Effective Shear Force, Ve, and Shear Resistance, Vn, Calculation

-500

0

500

1000

1500

2000

0 500 1000 1500 2000 2500 3000

Axi

al L

oad

(ki

p)

Moment (kip-in)

COL 1: P-M Interaction Diagram

Mu,Pu

Mn, Pn

ΦMn, ΦPn



o

o

o

(This is positive because the column is in compression)

Minimum Spacing for Low Demand

Spacing for column shear (ACI 318-11 §7.10.5.2)

Need to check Pu ≥ 0.35Ag f’c

Is 296 ≥ 0.35 ?

296 K < 358 K

No r qu r

Stirrup and Spacing Selection



The calculations for other columns in the system (C2 & C3) can be found in Appendix A.

Table 3.14: Column Schedule for two levels

COLUMN SCHEDULE

MARK DETAIL SIZE VERTICAL REINFORCEMENT

TIES

FIRST LEVEL C1 D1 16” x 16” 8 No. 6 No 4 @ 6” C2 D1 18” x 18” 8 No. 6 No 4 @ 4”

C3 D1 24” x 24” 8 No. 8 No 4 @ 5” BASEMENT

C1 D1 16” x 16” 8 No. 8 No 4 @ 4.5” C2 D1 18” x 18” 8 No. 8 No 4 @ 4.5” C3 D1 24” x 24” 8 No. 10 No 4 @ 6”

Campbell Hall “SHEAR WALLS” Pg. 1 out of 6 Date: 04/18/2014


SHEAR WALL DESIGN

First of all, we need to determine the lateral forces that need to be resisted by the shear wall during an

earthquake. We perform this analysis per ASCE 7-10 Seismic Design Criteria.

Seismic ground motion analysis

From USGS Detailed Report (http://earthquake.usgs.gov/designmaps):

The site is 37.82990N, 122.257030W

Per ASCE 7-10 Seismic Design Criteria section 11.4:

o Fundamental vibration period:

o

o

o

o

Calculation of building seismic weight, w

Loads at Roof Level

o

o

o

o

http://earthquake.usgs.gov/designmaps



o

o

Loads at the Office Level

o

o

o

o

o

o

Calculation of design base shear, V

o

But has to be less than:

And has to be greater than:

o

o

o



Calculation of lateral forces at roof and second level on each shear wall

o

o

(at second level)

o

(at roof level)

Select shear wall thickness, , such that the average shear stress on the wall does not exceed

o

Unfactored Dead Load PD and Live Load calculation on Shear Wall

Calculation of the amount of boundary element longitudinal reinforcement required for moment

strength at the wall base



Neutral axis, c, depth calculation

Depth of Boundary Element

o

o -

Boundary Element transverse steel calculation (consider Special Boundary Element)

o

o



Boundary Element transverse steel spacing calculation

Horizontal reinforcement for shear

5’

25’

4’’ spacing

6” spacing

15’



Required distributed vertical reinforcement

SHEAR WALL SCHEDULE

Table 4.1: Shear Wall Schedule

DESCRIPTION REINFORCEMENT

BOUNDARY ELEMENT LONGITUDINAL STEEL 12 No. 9 TRANSVERSE STEEL No. 4 @ 4”

CROSS-TIES No. 4 @ 4” INNER LOCATION OF SHEAR

WALL REQUIRED HORIZONTAL STEEL

FOR SHEAR No. 5 @ 12”

DISTRIBUTED VERTICAL STEEL No. 5 @ 12”

Campbell Hall “RETAINING WALL” Pg. 1 out of 5 Date: 02/14/2014


RETAINING WALL DESIGN

The retaining wall will surround the basement level in regions where there is not a shear wall.

The pressure from active Earth produces bending moment on the retaining wall. Using this

design moment, the wall is design as a one-way slab system to find the reinforcement required to

resist moment. The seismic forces generate a moment within the retaining wall. Using these

forces, this moment is calculated and reinforcement is selected to resist this moment. Finally,

temperature and shrinkage reinforcement is found for the retaining wall. The below calculations

adhere to ACI 318-11, ASCE 7, and ASTM reinforcement.

Givens (provided from criteria document and geotechnical report):

Allowable Moment, Mallow, Calculation:

Horizontal Earth Pressure Component

o

Seismic Pressure Component

o

o

30’

15’

16.5’ M C

T

F3

F1

F2

M T

C



o

o

o

o

Design as a one-way slab for reinforcement to resist earth pressure:

Initial Values:

Required Steel Area Calculation, As:

o

o

o

o

Table 4.2: Bar Size and Spacing

Bar Size Area (in2) Spacing (in)

5 0.31 8”

6 0.44 2”

Design like a beam for reinforcement o resist seismic lateral load:

Initial Value:

Required Steel Area Calculation, Aa:



o

Temperature and Shrinkage Steel:

o

o

o

Table 4.3: Bar Size and Spacing for Temperature and Shrinkage

Bar Size Area (in2) Spacing (in)

3 0.11 6”

4 0.2 ”

5 0.31 7”

Table 4.4: Retaining Wall Schedule

DESCRIPTION REINFORCEMENT

MAIN STEEL TO RESIST EARTH

PRESSURE

No. 6 @ 2”

MOMENT SEEL TO RESIST SEISMIC

LATERAL LOAD

9 No. 11 (IN 3 LAYERS)

TEMPERATURE AND SHRINKAGE STEEL No. 4 @ ”

3 layers of 9 No. 11

bars



2F

F

1.5F

.5F1 1.5F

T

T C

C

C

3F

97.5 F

1.5F

48.5F

1.5F

48.5F

1.5F

1.5F

T T C



In order to calculate reinforcement required for the retaining wall, the following steps are taken.

The wall has an axial force and a bending moment, therefore, it has to be designed like a column.

The axial force on the wall comes from the lateral forces. Since there are two retaining walls, the

axial force on each wall is equal to half of the sum of the two forces. P = 0.5 (2F + F). The

moment on the wall is also caused by the two lateral forces. M = 2F (37.5) + F (22.5). The wall

on the right has a compressive axial force, whereas the wall on the left has a tensile axial force.

Campbell Hall “DIAPHRAGM” Pg. 2 out of 2 Date: 04/30/2014


DIAPHRAGM DESIGN (ROOF LEVEL)

Loading calculation

Chord reinforcement

o

o

Collector

VU

VU’

11’

39’

25’

Section

cut

25’

VU’

Campbell Hall “DIAPHRAGM” Pg. 2 out of 2 Date: 04/30/2014


If Earthquake hits in the opposite direction

o

o

Compression check

o

(let s = 4 inches)

o

Note: We have to put the collectors and chord reinforcement all around the building to

anticipate earthquake coming in the other direction. Since collectors can also serve as chord

reinforcement if the earthquake hits in the other direction, and since the amount of collectors

needed is greater than the chord, we will place just the collectors all around the building.

Shear friction

We put all the collectors within the diaphragm region, so all the applied lateral load will be the shear

force in the diaphragm.

o

Pick 50 No. 6

Steel ratio check:

Campbell Hall “DIAPHRAGM DESIGN” Pg. 1 out of 3 Date: 03/30/2014


DIAPHRAGM DESIGN (1ST FLOOR)

Loading calculation

Chord reinforcement

o

o

Collector

If Earthquake hits in the opposite direction

VU

VU’

11’

12’

39’

25’ VU”

13’

Section

cut



o

o

Compression check

o

(let s = 3.5 inches)

o

Note: We have to put the collectors and chord reinforcement all around the building to

anticipate earthquake coming in the other direction. Since collectors can also serve as chord

reinforcement if the earthquake hits in the other direction, and since the amount of collectors

needed is greater than the chord, we will place just the collectors all around the building.

Shear friction

Since we put 6 out of 15 collector bars within the diaphragm, we have the shear force =

there.

Similarly, since we put 9 collectors in the retaining (basement) wall, we have the shear force =

there.

o

For the retaining wall region that has shear wall on top of it, the distributed vertical steel from shear

wall can serve as shear friction as well. In our case, we have 44 No. 5 at 12” as the distributed vertical

reinforcement. Hence, As = 44 * 0.31 = 13.64 inches2 ≥ 13.55 inches2 (OK).

For the retaining wall region that has no shear wall on top of it, the vertical moment steel from retaining

wall can serve as shear friction as well. In our case, we have No. 6 at 12” as the vertical moment steel

from retaining wall. This should be good for the shear friction because before we have No. 5 at 12” and

it is sufficient



For diaphragm shear friction design:

o

o =2.5

Pick 33 No. 6

Steel ratio check:

Campbell Hall “FOUNDATIONS” Pg. 1 out of 3 Date: 04/27/2014


STRIP FOOTING DESIGN

First, the width of the footing is determined based on dead and live loads. Then the footing is checked

for one way shear. The moment is used to determine the reinforcement required for the footing. The

development length of the bars and dowels is also checked. Finally, the length of the lap splice is

determined.

Analysis for loads

Dead Load

Load over 3 levels

Footing width calculation

One way shear check

o

o



o

o

o

∴ The footing width is adequate

Moment design

o

o

o

o

o

∴

Moment steel development length check

Provided Development Length Calculation

o

Required Development Length



o

o

Dowel placement

Using No. 5 bars, and check if the development length is sufficient

-

o

o

o

o

∴

STRIP FOOTING SCHEDULE

Table 4.2: Strip Footing Schedule

MARK WIDTH DEPTH REINFORCEMENT DOWELS

F4 5’ 1’ 6” No. 4 @ 12” 8 No. 5



SPREAD FOOTING DESIGN

Spread Footing for Column 1:

First, the dimensions of the footing are determined based on dead and live loads that were calculated

earlier and are included in the appendix. Then the footing is checked for one way shear and two way

shear. The moment is used to determine the reinforcement required for the footing. Bearing resistance,

the development length of the bars and dowels are also checked. Finally, the length of the lap splice is

determined.

C1 = 8” (square column)

D = 142.5 K

L= 78.13 K

Assumptions

Two cases are examined. In the first case, only dead load (D)

is considered and in the second case, dead and live loads

(D+L) are considered.

Case 1:

(recommended by Kutt and Filit)



Case 2:

(recommended by Kutt and Filit)

∴ A footing plan dimension of 8’x8’ is used.

Shear Check

One-way Shear:

First, one way shear is considered. In this case, the critical section is a distance d away from the column.

o

o

λ=1.00

-

o



o

∴These plan dimensions are adequate

Two-Way Shear:

o =0.75

o

As the column is positioned central to the footing =40

This will not dominate because it is a square footing & βc =1

Or



Or

∴

o

o

o

∴The current dimensions are sufficient

Moment Design

o

o

o

o

o

o

o

∴Use 6 No. 5 bars in both directions



Development Length

o

o

o

o

o λ=1

o

37” > 23.7”

∴ The length provided is sufficient

Dowel Placement

Using #8 dowel

b – the straight part of the dowel.

o

o



”

∴The provided is currently satisfactory and b=21.75”

Distance from Foundation Level to Lap Splice, a, Calculation

a = max:

∴ a=30” – this is the length of the lap splice

Bearing Resistance

o

o

∴

∴This footing size is adequate

Table 4.1: Spread Footing Schedule

Campbell Hall “APPENDIX A”


APPENDIX A

Column 2:


Mu = 763.7 k-in

Initial Values








design for the column. A rectangular column with 8 No.8 bars is inputted. The type of steel is specified

as A 615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain

is 0.003.

Figure A.1: P-M Interaction Diagram

As it can be seen in Figure A.1, the point Pu, Mu is inside of the curve, which means that the


Column Shear



-500

0

500

1000

1500

2000

2500

0 500 1000 1500 2000 2500 3000 3500 4000

Axi

al L

oad

(ki

p)

Moment (kip-in)


Mu, Pu

Mn, Pn

ΦMn, ΦPn



o

o

o





Is 574 ≥ 0.35 ?

574 K > 454K

Special requirement



o

o


Column 3:


Mu = 1445.9 K-in

Initial Values








design for the column. A rectangular column with 8 No.8 bars is inputted. The type of steel is specified

as A 615 Gr 60. A single No 4 hoop with two ties is used. Concrete cover is 1.5”, f’c = 4 ksi and the strain

is 0.003.



Figure A.2: P-M Interaction Diagram

As it can be seen in Figure A.2, the point Pu, Mu is inside of the curve, which means that the


Column Shear



o

o

o

-1000

-500

0

500

1000

1500

2000

2500

3000

3500

4000

0 2000 4000 6000 8000 10000

Axi

al L

oad

(ki

p)

Moment (kip-in)


Mu, Pu

Mn, Pn

ΦMn, ΦPn







Is 1112 ≥ 0.35 ?

1112K >806k

S r qu r

o



o


LOADING BREAKDOWN:

ROOF DEAD LOAD LIVE LOAD

SW 150 pcf Garden 100 psf

HVAC 15 psf

Light 5 psf

LEVEL 2 DEAD LOAD LIVE LOAD

SW 150 pcf Office 50 psf

HVAC 15 psf

Light 5 psf

Cladd 15 psf

LEVEL 1 DEAD LOAD LIVE LOAD

SW 150 pcf Classes 100 psf

HVAC 15 psf

Light 5 psf

Cladd 15 psf

BASEMENT DEAD LOAD LIVE LOAD

SW 150 pcf Shelf 300 psf

Cladd 15 psf

ROOF:

Total

ID DL (psf) LL (psf) DL (lb) LL (lb)1.4 DL

(psf)

1.2 DL

(psf)

1.6 LL

(psf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Slab S1 120 100 0 0 168 144 160 0 0 0 304

S2 120 100 0 0 168 144 160 0 0 0 304

S3 120 100 0 0 168 144 160 0 0 0 304

ID Width (in)Depth

(in)DL (plf) LL (plf) DL (lb) LL (lb)

1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Beam B1 12 24 950 625 0 0 1330 1140 1000 0 0 0 2140

B2 12 24 1700 1250 0 0 2380 2040 2000 0 0 0 4040

B3 12 24 1700 1250 0 0 2380 2040 2000 0 0 0 4040

ID Width (in)Depth


1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Girder G1 12 24 200 0 21250 15625 280 240 0 29750 25500 25000

G2 12 24 200 0 42500 31250 280 240 0 59500 51000 50000

ROOF:

ID

Design

Moment

(k-ft)

Design

Shear

(kips)

Slab S1

S2

S3

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Beam B1 12 24

B2 12 24

B3 12 24

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

8

8

8

8

8

8

Thickness (in)

Factored Concentrated LoadDistributed Concentrated Factored Distributed Load

Thickness (in)

Girder G1 12 24

G2 12 24

LEVEL 2:

Total


(psf)

1.2 DL

(psf)

1.6 LL

(psf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Slab S1 120 50 0 0 168 144 80 0 0 0 224

S2 120 50 0 0 168 144 80 0 0 0 224

S3 120 50 0 0 168 144 80 0 0 0 224

ID Width (in)Depth


1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Beam B1 12 24 1175 312.5 0 0 1645 1410 500 0 0 0 1910

B2 12 24 1700 625 0 0 2380 2040 1000 0 0 0 3040

B3 12 24 1700 625 0 0 2380 2040 1000 0 0 0 3040

ID Width (in)Depth


1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Girder G1 12 24 200 0 21250 7812.5 280 240 0 29750 25500 12500

G2 12 24 200 0 42500 15625 280 240 0 59500 51000 25000

LEVEL 2:

ID

Design

Moment

(k-ft)

Design

Shear

(kips)

Slab S1

S2

S3

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Beam B1 12 24

B2 12 24

B3 12 24

8

8

8

Thickness (in)

Thickness (in)

8

8

8

Concentrated Factored Distributed Load Factored Concentrated LoadDistributed

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Girder G1 12 24

G2 12 24

LEVEL 1:

Total


(psf)

1.2 DL

(psf)

1.6 LL

(psf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Slab S1 120 100 0 0 168 144 160 0 0 0 304

S2 120 100 0 0 168 144 160 0 0 0 304

S3 120 100 0 0 168 144 160 0 0 0 304

ID Width (in)Depth


1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Beam B1 12 24 1175 625 0 0 1645 1410 1000 0 0 0 2410

B2 12 24 1700 1250 0 0 2380 2040 2000 0 0 0 4040

B3 12 24 1700 1250 0 0 2380 2040 2000 0 0 0 4040

ID Width (in)Depth


1.4 DL

(plf)

1.2 DL

(plf)

1.6 LL

(plf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Girder G1 12 24 200 0 21250 15625 280 240 0 29750 25500 25000

G2 12 24 200 0 42500 31250 280 240 0 59500 51000 50000

LEVEL 1:

ID

Design

Moment

(k-ft)

Design

Shear

(kips)

Slab S1

S2

S3

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Beam B1 12 24

8

8

8

8

8

Factored Distributed Load Factored Concentrated Load

Thickness (in)

Thickness (in)

8

Distributed Concentrated

B2 12 24

B3 12 24

ID Width (in)Depth

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Girder G1 12 24 -191.091 183.678

G2 12 24 -367.134 359.472

BASEMENT:

Total


(psf)

1.2 DL

(psf)

1.6 LL

(psf)

1.4 DL

(lb)1.2 DL (lb)

1.6 LL

(lb)

Factored

Load

Slab S1 100 300 0 0 140 120 480 0 0 0 600

BASEMENT:

IDThickness

(in)

Design

Moment

(k-ft)

Design

Shear

(kips)

Slab S1

SLAB BEAM Factored Beam/ Girder: 1/2 SW (lb)

LEVEL 2:

IDTriburary

Area (ft2)

Factored

Load (psf)

Slab

Point

Load

(kip)

Factored

Beam/

Girder

(kip)

No. of

beams

No. of

beams

x load

No. of

girders

No. of

girders x

load

Total

Load on

Col (kip)

Column 1 156.25 220 34.375 2.5 1 2.5 1 2.5 39.375

2 312.5 220 68.75 2.5 2 5 1 2.5 76.25

3 625 220 137.5 2.5 2 5 2 5 147.5

LEVEL 1:

Factored Distributed Load Factored Concentrated Load

8

Distributed Concentrated

Thickness (in)

8

IDTriburary

Area (ft2)

Factored

Load (psf)

Slab

Point

Load

(kip)

Factored

Beam/

Girder

(kip)

No. of

beams

No. of

beams

x load

No. of

girders

No. of

girders x

load

Load

from

Level 2

(kip)

Total

Load on

Col (kip)

Column 1 156.25 170 26.5625 2.5 1 2.5 1 2.5 39.375 70.9375

2 312.5 170 53.125 2.5 2 5 1 2.5 76.25 136.875

3 625 170 106.25 2.5 2 5 2 5 147.5 263.75

BASEMENT:

IDTriburary

Area (ft2)

Factored

Load (psf)

Slab

Point

Load

(kip)

Factored

Beam/

Girder

(kip)

No. of

beams

No. of

beams

x load

No. of

girders

No. of

girders x

load

Load

from

Level 2

(kip)

Load

from

Level 1

(kip)

Total Load

on Col

(kip)

Column 1 156.25 220 34.375 2.5 1 2.5 1 2.5 39.375 70.9375 220.625

2 312.5 220 68.75 2.5 2 5 1 2.5 76.25 136.875 426.25

3 625 220 137.5 2.5 2 5 2 5 147.5 263.75 822.5

Beams and Girders Moment and

Shear Calculation

Center

B2 - MAX

M1 M2 M3 M4 M5 M6 M7 M8 M9

0 196.47 -271.29 92.38 -182.49 92.38 -271.29 92.38 0 MAX-MAX-MAX-MAX

B2 - MAX-MIN

0 222.13 -204.58 -10.57 -137.62 147.33 -204.58 74.65 0 MAX-MIN-MAX-MIN

B2 -

0 74.65 -204.58 147.33 -137.62 -10.57 -204.58 222.13 0 MIN-MAX-MIN-MAX

B2

0 192.22 -282.35 111.43 -147.62 28.73 -128.69 104.11 0 MAX-MAX-MIN-MIN

B2

0 190.14 -287.74 120.7 -115.79 1.27 -209.97 220.06 0 MAX-MAX-MIN-MAX

0 213.73 -226.42 35.59 -49.08 35.59 226.42 213.73 0 MAS-MIN-MIN-MAX

0 82.60 -182.75 113.07 -226.16 113.07 -182.75 82.6 0 MIN-MAX-MAX-MIN

0 215.8 -221.03 24.23 -70.93 56.22 -143.26 97.78 0 MAX-MIN-MIN-MIN

0 76.3 -199.2 138.05 -159.46 21.55 -121.43 106.18 0 MIN-MAX-MIN-MIN

0 106.18 -21.43 21.55 -159.46 138.05 -199.2 76.3 0 MIN-MIN-MAX-MIN

Critical Case

0 222.13 -287.74 147.33 -226.16 147.33 -271.29 222.13 0

B1

0 117.73 162.56 55.358 109.35 -189.53 107.36 0 MAX

0 123.36 -140.12 26.15 -42.65 -189.53 107.36 0 MAX-MIN-MAX

0 69.42 -95.85 32.54 -64.48 -189.53 107.36 0 MIN-MAX-MAX

Critical Case

0 123.36 -140.12 55.358 -64.48 -189.53 107.36 0

B3

Maximum Positive Moment 129.84 Combo 6

Maximum Negative Moment -235.92 Combo 6

Shear

B1 -24.86 36.07/-21.85 14.06 -38.04 22.88

B2 -42.65 59.02/-28.52 23.16/-48.16 53.52/-34.02 17.65

-39.33 62.35/-57.72 43.96/-22.07 29.16/-59.24 42.34

FOOTINGS Choosen OKAY IF BELOW:

Changes demand < capacity

D+L - unfactored Checks - demand (ex: Vu)

A_trib_wall 625 ft2 Checks - capacity (ex: ΦVn)

SLAB (psf) trib (ft)

D_slab

(plf)

factor for

B + G

D_slab per

level (plf) No. slabs

D_slab

total

Wall sw

(plf)

D total

(klf)

DEAD 120 12.5 1500 1.2 1800 3 5400 10125 15.525

3 levels (psf) trib (ft) L total (klf)

LIVE 250 12.5 3.125

COL 1

wall thickness 18 inches 1.5 ft

w D 15.525 ksf

w D + L 18.65 ksf

q_allow (D) 3.5 ksf

q_allow (D + L) 4.5 ksf

ρ RC 0.15 kcf

ρ soil 0.11 kcf

H 24 inches 2 ft

h 18 inches 1.5 ft

H + h = amount below grade

Dead + Live (1' strip)

q = (D + L) / ab + Hρ_soil + hρ_RC ≤ q_allow A = 4.60 ft

q D (klf) D + L (klf)Hρ_soil

(ksf)

hρ_RC

(ksf)

q_allow

(D) (ksf)

q_allow (D

+ L) (ksf)

15.525 18.65 0.22 0.225 3.5 4.5 » A = 5.08 ft

a = 5.00 ft

60 in

Dead (1' strip)

a

b

Bearing Pressure -> Shear

One way shear

f'c 4000 psi 4 ksi Vn = Vc + Vs, Vs = 0

Φ 0.75 Vc = 2*sqrt(f'c)*bw*d*λ

λ 1 Vu = q_u,net * l_b*a

concrete cover 3 inches kips

dia - No. 8 bar 1 inches Vc 22.01 kips

d 14.5 inches ΦVc 16.51 kips

c + d 32.5 inches Vu 2.56 kips

q_u,net = Pu/ab 4.726 ksf 0.032819 ksi

l_b 0.54166667 ft 6.5 inches

Moment

Φ 0.9 M_u2 86.84 k-in 7.236688 k-ft

q_u,net 4.726 ksf 0.0328 ksi Mn 96.49 k-in

l_b 21 inches 1.75 ft Mn = As fy d (1 - 0.59*(As/bd)*(fy/f'c))

fy 60 ksi As = (-b +/- sqrt(b^2-4ac))/(2a)

f'c 4 ksi a = 0.59*fy^2/ (bf'c)

b = -fyd

c = Mn

a 8.85

b -870

c 96.4892 98.194 in^2

As 0.111

just As -> No bars

No Area

(in^2)

No 3 0.11

No 4 0.2

No 5 0.31

No 6 0.44

No 7 0.6

No 8 0.79

No 9 1

No 10 1.27

select No. 4 @ 12"

Check l_d

l_provided 18 inches

fy 60000 psi 60 ksi

f'c 4000 psi 4 ksi

≤ No 6 l_d = fy Ψt Ψe db / (25λ sqrt(f'c)) = 18.97367 inches

≥ No 7 l_d = fy Ψt Ψe db / (20 λ sqrt(f'c)) = 23.71708 inches

Ψt 1 bottom bars

Ψe 1

λ 1

db No. 4 0.5

Ψs 0.8 < No. 6

(cb + ktr)/db 2.5

long formula l_d = 3 fy Ψt Ψe Ψs db/ (40 λ sqrt(f'c) (cb+ktr)/db)) = 11.3842 Okay if l_provided > l_d

Placement of dowel steel

db, dowel 0.625 inches assume No 5

db, c wall (No 4) 0.5 inches

fy 60000 psi 60 ksi

f'c 4000 psi 4 ksi

cover 3 inches

compression lap splice

lsc = 0.0005 fy db,dowel = 18.75 inches

ldc = 0.02 fy db, col / sqrt(f'c) = 9.486833 inches

ldc = 0.0003 fy db,col = 9 inches

a = 19 inches

hook end

db No. 4 0.5

ldc = 0.0003 fy db,col = 11.25 inches

l_dc = 0.02 fy db / sqrt(f'c) = 11.85854 inches

b = l_dcprov= h-cover-2db-4db, dow = 12 inches Okay if l_dc provided > l_dc

Shear Walls max axial loading - when all spans loaded

Vu = Φ 10 √(f'c) Acw when wall has no reinforcement, use factor 10

Vu = Φ 5 √(f'c) Acw when wall has no reinforcement, use factor 5

Acw = b l, b = height, l = width

length, lw 25 ft

height, hw 30 ft

S_DS 1.643 g

S_D1 0.683 g

To 0.0831406 s

Ts 0.415703 s

hn 30 ft

Ct 0.02 period parameter from Table 12.8-2

x 0.75

T = Ta 0.2563722 s > To < T < Ts < TL

Sa = S_DS 1.643 g

Effective Seismic Weight

Roof

SLAB G + B C 1 C 2 C 3 Wall ∑ D (k)

No. 1 52 4 4 9 4

120 5 2 2.53125 4.5 28.125

D (k) 1200 260 8 10.125 40.5 112.5 1631.125

L (k) 1000

w2 (k) 1881.125

Level 2

SLAB G + B C 1 C 2 C 3 Wall ∑ D (k)

No. 1 52 4 4 9 4

120 5 4 5.0625 9 56.25

D (k) 1200 260 16 20.25 81 225 1802.25

L (k) 500

w1 (k) 1927.25

w (k) 3808.375

Seismic Base Shear

V = CsW

le 1 Risk II

R 5 special reinforced shear wall

Cs 0.3286 g

1.2817302 g

0.072292 g

V base 1251.432 k

redundancy factor ρ 1

torsion amp factor 1.3

V design 1626.8616 K

V wall = 1/2V 813.43082 k per wall

Acw = Vu / (0.6*5*sqrt(f'c))

Acw 6595.6259 in2

Acw = 25' x b in

b 21.98542 in

Choose 18 in

Check required wall length - don't need this!!

φvn by NIST 151.78933 psi 0.151789 ksi

A req on wall 5358.946 in2

L req 297.71922 in 24.80994 ft

Vu 9835.9484

Fx = Cvx V

k 1

h 15 ft

∑ wihi 85342.5 k-ft

Cvx2 0.661262

Cvx1 0.338738

Fx2 537.89087 k At roof

Fx1 275.53995 k At level 2

∑ F 813.43082 k

Pu - unfactored

A_trib_wall 625 ft2

Roof + Level 2 + Level 1

SLAB (psf) G + B (k) Wall (k) ∑ D (k)

No. 3 24 1

120 2.5 168.75

D (k) 225 60 168.75 453.75

L (k) 156.25

HOW MUCH STEEL DO YOU NEED IN BOUNDARY?

∑ M = 0

M base = Mu 20269.825 k-ft

M n 22522.028 k-ft 270264.3 k-in

lw 25 ft 300 in

0.1 lw 2.5 ft 30 in

0.2 lw 5 ft 60 in

0.4 lw 10 ft 120 in

0.6 lw 15 ft 180 in

0.8 lw 20 ft 240 in

As'' 8.1 in2

Ts'' = T l 486 k

Pu small 408.375 k

Ts 678.9139 k

As 11.315232 in2

No Area

(in^2)

No. of

bars

No. of

bars

No 3 0.11 102.8657 103

No 4 0.2 56.57616 57

No 5 0.31 36.50075 37

No 6 0.44 25.71644 26

No 7 0.6 18.85872 19

No 8 0.79 14.32308 15

No 9 1 11.31523 12 Choose 12 No. 9 bars

No 10 1.27 8.909631 9

No 11 1.56 7.253354 8

No 14 2.25 5.028992 6

HOW DEEP IS NEUTRAL AXIS?

Pu max 622.625 k

Pu / (Ag f'c) 0.0288252

from NIST graph c/lw = 0.08 -> c = 0.08 lw

c 2 ft 24 in

DEPTH OF B.E. ?

min depth of BE, l_be = max of below conditions

c/2 12 in

c - 0.1 lw -6 in

cover 1.5 in

max depth of BE 21 in

Choose l_be 20 in

CONFINEMENT REINFORCEMENT?

bc1, horiz 17 in

bc2, vert 15 in

spacing

bw 18 in

s ≤ 6 db, longit (no 9) 6.768 in

s ≤ bw/3 6 in <- controls

Choose 4 in

Choose 3 vert legs use bc1

Ash ≥ 0.09 s bc f'c / fy 0.408 in2

A per leg 0.136 in2 <- w/ 3 legs at No 4 bar

No Area

(in^2)

No. of

bars

No. of

bars

No 3 0.11 3.709091 4

No 4 0.2 2.04 3

No 5 0.31 1.316129 2

Choose 3 horiz leg use bc2

Ash ≥ 0.09 s bc f'c / fy 0.36 in2

A per leg 0.12 in2 <- w/ 3 legs at No 4 bar

No Area

(in^2)

No. of

bars

No. of

bars

No 3 0.11 1.090909 2

No 4 0.2 0.6 1

No 5 0.31 0.387097 1

WALL SHEAR - MIDDLE OF SHEAR WALL

Vu ≤ phi Vn

hw/ lw 1.2

α 3

pick No 3 for At middle horiz steel

Vn 1355.718 k

ρt 0.001022 < 0.0025

thus, ρt 0.0025

ρt = A t /(bw s) -> s 4.8888889 in Choose 5 in

REQUIRED DISTRIBUTED VERTICAL STEEL - MIDDLE OF SHEAR WALL

ρt 0.0025

ρl = 0.0025 + 0.5*(2.5 - hw/lw) (ρ t - 0.0025)

ρl 0.0025

Retaining Wall

σ a 30 pcf > w 792 plf

σ s 7h > w' 161.7 plf

length 16.5 ft

Mmax = M1 13.79981 k-ft soil

M2 5.502853 k-ft EQ

M1 + M2 (allow) 19.30266 k-ft 231.6319 k-in

Mn = Mu/ 0.9 21.4474 k-ft 257.3688 k-in

Design as one-way slab

thickness (l/20) 9.9 in use 14 in

d = 0.8h, h=1' 9.6 in

Mn = As fy d (1 - 0.59(Asfy/(bd*f'c)))

As = (-b +/- sqrt(b^2-4ac))/(2a)

a = 0.59*fy^2/ (bf'c)

b = -fyd

c = Mn

a 37.92857

b -576

c 257.3688

As 0.461 14.726 in^2

just As -> No bars

No Area

(in^2) s = 12*Ab/As

No 3 0.11 2.864564

No 4 0.2 5.208299

No 5 0.31 8.072864

No 6 0.44 11.45826

No 7 0.6 15.6249

No 8 0.79 20.57278

No 9 1 26.0415

No 10 1.27 33.0727

No 11 1.56 40.62473

No 14 2.25 58.59336

Choose No. 6 @ 12"

∑ M Q = - (15+16.5/2)*Fx1 - (30+16.5/2) *(Fx2) + M + F3*16.5/2= 0

F x2 537.8909 k

F x1 275.5399 k

F x3 813.4308 k

M 20269.83 k-ft 243237.9 k-in

0.5 M (b/c 2 side) 10134.91 k-ft 121619 k-in

Design like a beam

b 14 in

h 16.5 ft

d = 0.8*h 13.2 ft 158.4 in

Mn = As fy d (1 - 0.59(Asfy/(bd*f'c)))

As = (-b +/- sqrt(b^2-4ac))/(2a)

a = 0.59*fy^2/ (bf'c)

b = -fyd

c = Mn

a 37.92857

b -9504

c 121619

As 13.527 237.049 in^2

just As -> No bars

No Area

(in^2) No bars

No 3 0.11 122.9711 123

No 4 0.2 67.63412 68

No 5 0.31 43.63491 44

No 6 0.44 30.74278 31

No 7 0.6 22.54471 23

No 8 0.79 17.12256 18

No 9 1 13.52682 14

No 10 1.27 10.65104 11

No 11 1.56 8.671041 9

No 14 2.25 6.011922 7

Choose 14 No. 9

T & S

ρ TS 0.0025

b 14 in

A_TS = sh* ρ TS

A TS = 0.035 s

s ≤ A TS/ 0.035

No Area

(in^2) s (in)

No 3 0.11 6.285714 6

No 4 0.2 11.42857 11

No 5 0.31 17.71429 17

Choose 2 No 4 at 10 inches

FOOTINGS OKAY IF BELOW: Choosen

demand < capacity Changes

COL 1 Checks - demand (ex: Vu)

C1 16 inches 1.333333 ft Checks - capacity (ex: ΦVn)

C2 16 inches 1.333333 ft

Pu 296 kips

D 142.5 kips

D + L 220.625 kips

q_allow (D) 3.5 ksf

q_allow (D + L) 4.5 ksf

ρ RC 0.15 kcf

ρ soil 0.11 kcf

H 24 inches 2 ft

h 30 inches 2.5 ft

H + h = amount below grade Dead + Live

A = 56.50 ft2

q = (D + L) / ab + Hρ_soil + hρ_RC ≤ q_allow a = b = 7.52 ft

q D (k) D + L (k)Hρ_soil

(ksf)

hρ_RC

(ksf)

q_allow

(D) (ksf)

q_allow (D

+ L) (ksf)Dead

142.5 220.625 0.22 0.375 3.5 4.5 » A = 49.05 ft2

a = b = 7.00 ft

a = b= 8 ft

96 in

Bearing Pressure -> Shear

One way shear

f'c 4000 psi 4 ksi Vn = Vc + Vs, Vs = 0

Φ 0.75 Vc = 2*sqrt(f'c)*bw*d*λ

λ 1 Vu = q_u,net * l_b*a

concrete cover 3 inches

dia - No. 8 bar 1 inches Vc 315.72 kips

d 26 inches ΦVc 236.79 kips

c + d 42 inches Vu 43.17 kips

q_u,net 4.625 ksf 0.0321 ksi

a

b

c1

c2

l_b 14 inches

Two way shear Vu 239.34 kips

bo 168 inches Vc ≤ 1105.03 kips

Φ 0.75 βc 1657.54 kips

α_s 40 center α_s 2262.67 kips

βc 1 ΦVc 828.77 kips

Moment

Φ 0.9 M_u2 2466.67 k-in

q_u,net 4.625 ksf 0.0321 ksi Mn 2740.74 k-in

l_b 40 inches Mn = As fy d (1 - 0.59*(As/bd)*(fy/f'c))

fy 60 ksi As = (-b +/- sqrt(b^2-4ac))/(2a)

f'c 4 ksi a = 0.59*fy^2/ (bf'c)

b = -fyd

c = Mn

a 5.53125

b -1560

c 2740.741

As 1.768 280.266 in^2

just As -> No bars

No Area

(in^2)

No. of

bars

No 3 0.11 16.07243

No 4 0.2 8.839839

No 5 0.31 5.703122

No 6 0.44 4.018109

No 7 0.6 2.946613

No 8 0.79 2.237934

No 9 1 1.767968

No 10 1.27 1.392101

select 6 No. 5

Check l_d

l_provided 37 inches

fy 60000 psi 60 ksi

f'c 4000 psi 4 ksi

≤ No 6 l_d = fy Ψt Ψe db / (25λ sqrt(f'c)) = 23.71708 inches Okay if l_provided > l_d

≥ No 7 l_d = fy Ψt Ψe db / (20 λ sqrt(f'c)) = 29.64635 inches

Ψt 1 bottom bars

Ψe 1

λ 1

db No. 5 0.625

Placement of dowel steel

db, dowel 1 inches assume No 8

db, col (No 8) 1 inches

fy 60000 psi 60 ksi

f'c 4000 psi 4 ksi

cover 3 inches

compression lap splice

lsc = 0.0005 fy db,dowel = 30 inches

ldc = 0.02 fy db, col / sqrt(f'c) = 18.97367 inches

ldc = 0.0003 fy db,col = 18 inches

a = 30 inches

hook end

db No. 5 0.625

l_dc = 0.02 fy db / sqrt(f'c) = 18.97367 inches

l_dcprov= h-cover-2db-4db, dow = 21.75 inches Okay if l_dc provided > l_dc

Check Bearing

Φ 0.65 Pn = 0.85f'cAc*sqrt(A2/A1) + Asfy

y = -2x+2h 60 when x = 0 Pn 2120 kips

fy 60000 psi 60 ksi ΦPn 1378 kips > 296 Pu

f'c 4000 psi 4 ksi Okay

Ac = A col 256 in^2

A1 (col) 256 in^2

A2 (bearing) 18496 in^2

sqrt(A2/A1) ≤ 2 8.5 ≤ 2

Adowel = As 8 No 8 (8 bars b/c its that many in col)

6.32 in^2

Documents

Structural Design Project